Categories > vectors-m > Modulus of vector > If ABCD is a parallelogram, $\overrightarrow{AB}=2\,\mathbf{i}+4\,\mathbf{j}-5\,\mathbf{k}$ and $\overrightarrow{AD}=\,\mathbf{i}+2\,\mathbf{j}+3\,\mathbf{k},$ then the unit vector in the direction of BD is

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Question: If ABCD is a parallelogram, $\overrightarrow{AB}=2\,\mathbf{i}+4\,\mathbf{j}-5\,\mathbf{k}$ and $\overrightarrow{AD}=\,\mathbf{i}+2\,\mathbf{j}+3\,\mathbf{k},$ then the unit vector in the direction of BD is


1) $\frac{1}{\sqrt{69}}\,(\mathbf{i}+2\mathbf{j}-8\mathbf{k})$
2) $\frac{1}{69}\,(\mathbf{i}+2\mathbf{j}-8\,\mathbf{k})$
3) $\frac{1}{\sqrt{69}}\,(-\mathbf{i}-2\mathbf{j}+8\mathbf{k})$
4) $\frac{1}{69}\,(-\mathbf{i}-2\mathbf{j}+8\,\mathbf{k})$
Solution: Explanation: No Explanation
Modulus of vector Algebra of vectors

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